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3 years ago

Please answer as soon as possible gggggg

Mathematics

2 answers:

attashe74 [19]3 years ago

7 0

3/14x is the answer to this question

fiasKO [112]3 years ago

3 0

Answer:

X 3/14

Step-by-step explanation:

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The measures of the exterior angles of an octagon are x°, 2x", 3x°, 4x °, 6x°, 7xº,

Neporo4naja [7]

Answer:

216

Step-by-step explanation:

Total of octagon amount is 1080, so set it equal.

x + 2x + 3x + 4x + 6x + 7x + 8x + 9x = 1080

45x = 1080

/45 /45

x = 24

Then plug in for largest number.

9 (24) = 216

8 0

3 years ago

Two runners in a marathon determine that the angles of elevation of a news helicopter are 39.5 degrees and 42 degrees. If the he

Liula [17]

See the attached figure to understand the problem

we know that
tan 42°=1950/BO---------> BO=1950/tan 42°-------> BO=2165.69 ft

tan 39.5°=1950/AO---------> AO=1950/tan 39.5°-------> AO=2365.54 ft

AB=AO-BO--------> AB=2365.54 ft-2165.69 ft-------> AB=199.85 ft

the answer is
The runners are 199.85 feet apart

8 0

3 years ago

Which graph is the solution to x < 6

uysha [10]

Answer:

Step-by-step explanation:

hope this helps

7 0

3 years ago

Read 2 more answers

Solve the system of equations by using the inverse of the coefficient matrix of the equivalent matrix equation.

Alinara [238K]

The solution for the given system of equations x + 8y = -37, 4x + 8y = -52 is $\left[\begin{array}{ccc}&5&\\\\&4&\end{array}\right]$

Given,

System of equations as,

x + 8y = -37

4x + 8y = -52

We have to solve this by using the inverse of coefficient matrix of the equivalent matrix equation.

That is,

$A=\left[\begin{array}{ccc}a&&b\\\\c&&d\end{array}\right]$

$A^{-1} =\frac{1}{ad -bc} \left[\begin{array}{ccc}d&&-b\\\\-c&&a\end{array}\right]$

Now we can solve the equations.

Here we have,

x + 8y = -37

4x + 8y = -52

Now in matrix form,

$\left[\begin{array}{ccc}1&&8\\\\4&&8\end{array}\right]$ $\left[\begin{array}{ccc}&x&\\\\&y&\end{array}\right]$ $=\left[\begin{array}{ccc}&-37&\\\\&-52&\end{array}\right]$

A X B

We know that,

$A^{-1} =\frac{1}{ad -bc} \left[\begin{array}{ccc}d&&-b\\\\-c&&a\end{array}\right]$

Therefore,

$A^{-1} = \frac{1}{(1X8)-(4X8)} \left[\begin{array}{ccc}8&&-8\\\\-4&&1\end{array}\right]$

$=\frac{1}{8-32} \left[\begin{array}{ccc}8&&-8\\\\-4&&1\end{array}\right]$

$=\frac{1}{-24} \left[\begin{array}{ccc}8&&-8\\\\-4&&1\end{array}\right]$

$=\left[\begin{array}{ccc}\frac{-8}{24} &&\frac{8}{24} \\\\\frac{4}{24} &&\frac{-1}{24} \end{array}\right]$

Then,

$\left[\begin{array}{ccc}&x&\\\\&y&\end{array}\right] =$ $=\left[\begin{array}{ccc}\frac{-8}{24} &&\frac{8}{24} \\\\\frac{4}{24} &&\frac{-1}{24} \end{array}\right]$ $\left[\begin{array}{ccc}&\frac{37}{52} &\\\\\end{array}\right]$